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駭客數學指數5題
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Re: 駭客數學指數5題
第 1 題
a = log5 = 1 - log2
1/log2 = 1/(1 - a)
b = log6/log2 = (log2 + log3)/log2 = 1 + (log3/log2)
log3/log2 = b - 1
所求 = log0.12/log4 = (log12 - 2)/2log2 = (2log2 + log3 - 2)/2log2 = 1 + (1/2)(log3/log2) - 1/log2 = 1 + (1/2)(b - 1) - 1/(1 - a) = (ab + a - b + 1)/(2a - 2)
第 2 題
log(7^20) = 20log7 = 16.902
7^20 是 17 位數
log[7^20/10^16] = 0.902
log7 < log[7^20/10^16] < log8
7 < 7^20/10^16 < 8
7 * 10^16 < 7^20 < 8 * 10^16
故所求 = 7
第 3 題
以下 log 均以 a 為底,0 < a < 1
log(2x + 1) + logx ≧ 0
log(2x^2 + x) ≧ 0
2x^2 + x ≦ 1
-1 ≦ x ≦ 1/2
但由 logx 知 x 必大於 0
故 0 < x ≦ 1/2
第 4 題
以下 log 均以 2 為底
(logx)^2 + logx - 2 ≦ 0
(logx + 2)(logx - 1) ≦ 0
-2 ≦ logx ≦ 1
1/4 ≦ x ≦ 2
第 5 題
令 y = 9^log5 (以 3 為底)
logy (以 3 為底) = log5 * log9 (以 3 為底) = 2log5 (以 3 為底) = log25 (以 3 為底)
y = 25
log1 (以 3 為底) = 0
log8 (以 4 為底) = log2^3 (以 2^2 為底) = 3/2
log7 (以 1/7 為底) = -1
......
a = log5 = 1 - log2
1/log2 = 1/(1 - a)
b = log6/log2 = (log2 + log3)/log2 = 1 + (log3/log2)
log3/log2 = b - 1
所求 = log0.12/log4 = (log12 - 2)/2log2 = (2log2 + log3 - 2)/2log2 = 1 + (1/2)(log3/log2) - 1/log2 = 1 + (1/2)(b - 1) - 1/(1 - a) = (ab + a - b + 1)/(2a - 2)
第 2 題
log(7^20) = 20log7 = 16.902
7^20 是 17 位數
log[7^20/10^16] = 0.902
log7 < log[7^20/10^16] < log8
7 < 7^20/10^16 < 8
7 * 10^16 < 7^20 < 8 * 10^16
故所求 = 7
第 3 題
以下 log 均以 a 為底,0 < a < 1
log(2x + 1) + logx ≧ 0
log(2x^2 + x) ≧ 0
2x^2 + x ≦ 1
-1 ≦ x ≦ 1/2
但由 logx 知 x 必大於 0
故 0 < x ≦ 1/2
第 4 題
以下 log 均以 2 為底
(logx)^2 + logx - 2 ≦ 0
(logx + 2)(logx - 1) ≦ 0
-2 ≦ logx ≦ 1
1/4 ≦ x ≦ 2
第 5 題
令 y = 9^log5 (以 3 為底)
logy (以 3 為底) = log5 * log9 (以 3 為底) = 2log5 (以 3 為底) = log25 (以 3 為底)
y = 25
log1 (以 3 為底) = 0
log8 (以 4 為底) = log2^3 (以 2^2 為底) = 3/2
log7 (以 1/7 為底) = -1
......